Problem: Simplify the following expression and state the condition under which the simplification is valid. $k = \dfrac{6r^2 - 78r + 252}{-4r^2 + 48r - 144}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ k = \dfrac {6(r^2 - 13r + 42)} {-4(r^2 - 12r + 36)} $ $ k = -\dfrac{6}{4} \cdot \dfrac{r^2 - 13r + 42}{r^2 - 12r + 36} $ Simplify: $ k = - \dfrac{3}{2} \cdot \dfrac{r^2 - 13r + 42}{r^2 - 12r + 36}$ Next factor the numerator and denominator. $ k = - \dfrac{3}{2} \cdot \dfrac{(r - 6)(r - 7)}{(r - 6)(r - 6)}$ Assuming $r \neq 6$ , we can cancel the $r - 6$ $ k = - \dfrac{3}{2} \cdot \dfrac{r - 7}{r - 6}$ Therefore: $ k = \dfrac{ -3(r - 7)}{ 2(r - 6)}$, $r \neq 6$